1) The document derives an equation to calculate the velocity of an accelerating spaceship over time by considering infinitesimal changes in velocity and using the velocity addition formula. 2) This yields an equation showing velocity is equal to the speed of light multiplied by the hyperbolic tangent of acceleration times time over the speed of light. 3) The document then introduces the concept of rapidity, defined as the time integral of acceleration, and shows that rapidities add through simple addition while velocities add through a more complex formula.

1. Solution
Week 51 (9/1/03)
Accelerating spaceship
We will solve this problem by considering two nearby times and using the velocity-
addition formula,
v =
v1 + v2
1 + v1v2/c2
. (1)
Using the definition of the proper acceleration, a, we have (with v1 ≡ v(t) and
v2 ≡ a dt)
v(t + dt) =
v(t) + a dt
1 + v(t)a dt/c2
. (2)
Expanding both sides to first order in dt yields1
dv
dt
= a 1 −
v2
c2
. (3)
Separating variables and integrating gives, using 1/(1−z2) = 1/2(1−z)+1/2(1+z),
v
0
1
1 − v/c
+
1
1 + v/c
dv = 2a
t
0
dt. (4)
This yields ln ((1+v/c)/(1−v/c)) = 2at/c. Exponentiating, and solving for v, gives
v(t) = c
e2at/c − 1
e2at/c + 1
= c tanh(at/c). (5)
Note that for small a or small t (more precisely, for at/c 1), we obtain v(t) ≈ at,
as we should. And for at/c 1, we obtain v(t) ≈ c, as we should.
Remarks: If a happens to be a function of time, a(t), then we can’t move the a outside
the integral in eq. (4), so we instead end up with the general formula,
v(t) = c tanh
1
c
t
0
a(t) dt . (6)
If we define the rapidity, φ, by
φ(t) ≡
1
c
t
0
a(t) dt, (7)
then we have
v = c tanh φ ⇐⇒ tanh φ =
v
c
. (8)
Note that whereas v has c as a limiting value, φ can become arbitrarily large. The
φ associated with a given v is simply 1/mc times the time integral of the force (felt by
the astronaut) needed to bring the astronaut up to speed v. By applying a force for an
arbitrarily long time, we can make φ arbitrarily large.
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Equivalently, just take the derivative of (v + w)/(1 + vw/c2
) with respect to w, and then set
w = 0.
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2. The quantity φ is very useful because many expressions in relativity (which we’ll just
invoke here) take on a particularly nice form when written in terms of φ. Consider, for
example, the velocity-addition formula. Let β1 = tanh φ1 and β2 = tanh φ2. Then if we add
β1 and β2 using the velocity-addition formula, eq. (1), we obtain
β1 + β2
1 + β1β2
=
tanh φ1 + tanh φ2
1 + tanh φ1 tanh φ2
= tanh(φ1 + φ2), (9)
where we have used the addition formula for tanh φ (which can be proved by writing things
in terms of the exponentials e±φ
). Therefore, while the velocities add in the strange manner
of eq. (1), the rapidities add by standard addition.
The Lorentz transformation,
x
ct
=
γ γβ
γβ γ
x
ct
, (10)
also takes a nice form when written in terms of the rapidity. Note that γ can be written as
γ ≡
1
1 − β2
=
1
1 − tanh2
φ
= cosh φ, (11)
and so
γβ ≡
β
1 − β2
=
tanh φ
1 − tanh2
φ
= sinh φ. (12)
Therefore, the Lorentz transformation becomes
x
ct
=
cosh φ sinh φ
sinh φ cosh φ
x
ct
. (13)
This looks similar to a rotation in a plane, which is given by
x
y
=
cos θ − sin θ
sin θ cos θ
x
y
, (14)
except that we now have hyperbolic trig functions instead of the usual trig functions. The
fact that the invariant interval, s2
≡ c2
t2
− x2
, does not depend on the frame is clear from
eq. (13), because the cross terms in the squares cancel, and cosh2
φ−sinh2
φ = 1. (Compare
with the invariance of r2
≡ x2
+ y2
for rotations in a plane.)
Quantities associated with a Minkowski diagram also take a nice form when written in
terms of the rapidity. In particular, the angle between the axes of the two relevant frames
happens to be tan θ = β, where βc is the relative speed between the frames. But β = tanh φ,
so the angle between the axes is given by
tan θ = tanh φ. (15)
The integral a(t) dt (which is c times the rapidity) may be described as the naive,
incorrect speed. That is, it is the speed the astronaut might think he has, if he has his eyes
closed and knows nothing about the theory of relativity. (And indeed, his thinking would be
essentially correct for small speeds.) The quantity a(t) dt seems like a reasonably physical
thing, so if there is any justice in the world, a(t) dt = F(t) dt/m should have some
meaning. And indeed, although it doesn’t equal v, all you have to do to get v is take a tanh
and throw in some factors of c.
The fact that rapidities add via simple addition when using the velocity-addition for-
mula, as we saw in eq. (9), is evident from eq. (6). There is really nothing more going on
here than the fact that
t2
t0
a(t) dt =
t1
t0
a(t) dt +
t2
t1
a(t) dt. (16)
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3. To be explicit, let a force be applied from t0 to t1 that brings a mass up to speed
β1 = tanh φ1 = tanh(
t1
t0
a dt), and then let an additional force be applied from t1 to t2 that
adds on an additional speed of β2 = tanh φ2 = tanh(
t2
t1
a dt) (relative to the speed at t1).
Then the resulting speed may be looked at in two ways: (1) it is the result of relativistically
adding the speeds β1 = tanh φ1 and β2 = tanh φ2, and (2) it is the result of applying the
force from t0 to t2 (you get the same final speed, of course, whether or not you bother to
record the speed along the way at t1), which is β = tanh(
t2
t0
adt) = tanh(φ1 + φ2), where
the last equality comes from the obvious statement, eq. (16). Therefore, the relativistic
addition of tanh φ1 and tanh φ2 gives tanh(φ1 + φ2), as we wanted to show.
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